Chapter 21: Problem 8
If \(A B=3 i+j-k\) and \(A C=i-j+3 k .\) If the point \(P\) on the line segment \(B C\) is equidistant from \(A B\) and \(A C\), then \(A P\) is (A) \(2 i-k\) (B) \(i-2 \mathrm{k}\) (C) \(2 i+k\) (D) none of these
Short Answer
Expert verified
AP is \(2i - k\), option (A).
Step by step solution
01
Define Point P on Line Segment BC
Let's assume the coordinates of the point \( P \) on the line segment \( BC \) as \( B + t( C - B ) \), where \( 0 \leq t \leq 1 \). This is a parameterization where \( t \) is a fraction of the segment from \( B \) to \( C \).
02
Vector Definition for BC
First, compute the vector \( BC \). Since \( B \) and \( C \) lie on the vector line segments \( AB \) and \( AC \) respectively, we use the information: \( B = A + AB \) and \( C = A + AC \). Given the vectors \( AB = 3i + j - k \) and \( AC = i - j + 3k \), calculate \( BC = AC - AB \).
03
Calculate Vector BC
\[ BC = (i - j + 3k) - (3i + j - k) = (i - 3i) + (-j - j) + (3k + k) = -2i - 2j + 4k \]. This is the vector in the direction from \( B \) to \( C \).
04
Position of P on BC
The point \( P \) can be represented as: \[ P = B + t \cdot BC \]. Substitute \( B = 3i + j - k \) and \( BC = -2i - 2j + 4k \). \[ P = (3i + j - k) + t(-2i - 2j + 4k) = (3 - 2t)i + (1 - 2t)j + (-1 + 4t)k \].
05
Equidistance Condition
Since \( P \) is equidistant from both \( A \) via \( AB \) and \( AC \), set the magnitudes of \( AP \) from both directions equal. Calculate both resultant vectors from \( P \)'s position expressions and use equality condition.
06
Solve Distance Equations
Set magnitudes \( |AP_{viaAB}| = |AP_{viaAC}| \). Compute vectors \( AP \) using the coordinates of \( P \). Since \( t \) is fractional for equidistance, solve for \( A = P \) on segment, expressing in vector format \( AP = Ai + Bj + Ck \).
07
Simplify and Verify Result
On solving, you find \( t = 0.5 \) meaning equal division of segment equivalently, calculate \( AP = 2i - k \). Verify if equidistant point calculations hold.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parametric Equations
When dealing with line segments in space, parametric equations are a handy tool. Essentially, these equations allow us to express points on a line segment using a parameter, typically denoted as \( t \).
For instance, if you have a line segment between points \( B \) and \( C \), any point \( P \) on that segment can be defined by the equation:
For instance, if you have a line segment between points \( B \) and \( C \), any point \( P \) on that segment can be defined by the equation:
- \( P = B + t(C - B) \)
- When \( t = 0 \), \( P \) coincides with \( B \).
- When \( t = 1 \), \( P \) coincides with \( C \).
- In between, \( P \) simply moves along from \( B \) to \( C \).
Magnitude of Vectors
In vector algebra, one of the most fundamental operations is finding the magnitude—or length—of a vector. The magnitude of a vector \( \vec{v} = ai + bj + ck \) in a 3D space is calculated using the formula:
- \[ |\vec{v}| = \sqrt{a^2 + b^2 + c^2} \]
- \( |AP| = |PC| \)
Equidistant Points
An equidistant point is one that is the same distance from two separate points. In vector space, if a point \( P \) is equidistant from points \( A \) and \( C \), then it satisfies the condition:
To find such a point on a line segment, consider the position \( P = B + t(C - B) \) where \( t \) is chosen so that the distance from \( A \) to \( P \) equals the distance from \( P \) to \( C \). In the solved problem, this involves setting the magnitudes of vectors \( AP \) and \( PC \) equal, and solving for \( t \).
Write the equations for both distances, equate them, and, by solving, you'll find the needed parameter, \( t \), and thus, the coordinates of \( P \).
- \( |AP| = |PC| \)
To find such a point on a line segment, consider the position \( P = B + t(C - B) \) where \( t \) is chosen so that the distance from \( A \) to \( P \) equals the distance from \( P \) to \( C \). In the solved problem, this involves setting the magnitudes of vectors \( AP \) and \( PC \) equal, and solving for \( t \).
Write the equations for both distances, equate them, and, by solving, you'll find the needed parameter, \( t \), and thus, the coordinates of \( P \).
Line Segments in Space
Line segments in space are straight connections between two points represented by vectors. Understanding these segments involves grasping vector operations and parametric equations.
To define a segment between two points, such as point \( B \) and point \( C \), we use vector subtraction and addition. The vector representation \( BC = C - B \) gives us directionality and magnitude—a crucial component in navigation through 3D spaces.
To define a segment between two points, such as point \( B \) and point \( C \), we use vector subtraction and addition. The vector representation \( BC = C - B \) gives us directionality and magnitude—a crucial component in navigation through 3D spaces.
- The line segment is parameterized as \( P = B + t(C - B) \), where \( 0 \leq t \leq 1 \).